Answer
(a) $v = 7.79~m/s$
(b) The magnitude of acceleration is $a = 50.6~m/s^2$ and it is directed upward.
(c) $F_N = 4530~N$
$F_N = 6.16\times ~weight$
Work Step by Step
(a) We can find his speed $v$ when his feet first touch the ground.
$v^2 = v_0^2+2ay = 0+2ay$
$v = \sqrt{(2)(9.80~m/s^2)(3.10~m)}$
$v = 7.79~m/s$
(b) We can find the rate of deceleration as he slows down.
$a = \frac{v^2-v_0^2}{2y} = \frac{0-(7.79~m/s)^2}{(2)(0.60~m)}$
$a = -50.6~m/s^2$
The magnitude of acceleration is $a = 50.6~m/s^2$ and it is directed upward.
(c) $\sum F = F_N - mg$
$F_N - mg = ma$
$F_N = ma + mg = (75.0~kg)(50.6~m/s^2)+(75.0~kg)(9.80~m/s^2)$
$F_N = 4530~N$
We can express the average force $F_N$ as a multiple of his weight $mg$.
$F_N = \frac{4530~N}{(75.0~kg)(9.80~m/s^2)} = 6.16\times ~weight$
